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(this question were asked after studying line integrals)

1- Show that if $C$ is the graph of $y=f(x)$, $a \leq x \leq b,$ and if $F$ is a function of 2 variables defined on C, then $$\int_{C} F(x,y)dx = \int_{a}^{b} F(x, f(x))dx. $$

But I do not understand the meaning of the question, does it means that if I change the graph by its parametrization I will change the integration along the graph to the integration at the end points of the interval $[a,b]$, if so why is this correct?

Also could anyone give me a hint for the proof please?

Thanks!

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  • $\begingroup$ Are you sure there is not a Jacobian missing, e.g. $\sqrt{1+f'(x)^2}$ ? $\endgroup$ – Thomas Apr 18 at 7:09
  • $\begingroup$ @Thomas no there is not ..... this is exactly the question I have to answer. $\endgroup$ – Intuition Apr 18 at 7:11
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    $\begingroup$ Looks a bit weird to me but maybe I am not understanding. The left member when F(x,y)=1 should not be the length of C ? When parametrizing the curve by $t \rightarrow [t,f(t)]$ tutorial.math.lamar.edu/Classes/CalcII/ParaArcLength.aspx we would get a Jacobian more ? Maybe I am not getting the meaning of the let side... $\endgroup$ – Thomas Apr 18 at 7:19
  • $\begingroup$ May be the question contains a typo (which I am not sure of it because my professor have seen the question and did not say anything about it ..... or may be he\she was not concentrating at that moment) ..... I do not know @Thomas $\endgroup$ – Intuition Apr 18 at 7:29
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    $\begingroup$ Maybe I got the catch. I did not notice that you have a dx in the left member, which must be maybe intended as the integral of the differential form "F dx" over C. Before I interpreted the left integral as simply the integration of a scalar function over the manyfold C. $\endgroup$ – Thomas Apr 18 at 12:37
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$C$ is defined by $$C := \{ (x,y) \ | \ x\in[a,b], \ f(x) = y \}$$ so $$\int_{C} F(x,y)dx = \int_{\substack{x\in [a,b] \\ y = f(x)}} F(x,y)dx = \int_{x\in[a,b]} F(x,f(x))dx =\int_a^b F(x,f(x))dx$$ This just represents any function along $C$. For example the length of the curve would be $$\int_C \sqrt{dx^2 + dy^2} = \int_C\sqrt{1 + (\frac{dy}{dx})^2}dx = \int_a^b \sqrt{1 + (f'(x))^2}dx$$ Here $F(x,y) = \sqrt{1 + (\frac{dy}{dx})^2}$

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  • $\begingroup$ what about the proof? $\endgroup$ – Intuition Apr 18 at 10:46
  • $\begingroup$ I made an edit which should be enough justification. It could almost be used as a definition in my opinion. $\endgroup$ – Dayton Apr 18 at 10:56
  • $\begingroup$ Thank you so much for you great effort. $\endgroup$ – Intuition Apr 18 at 10:59
  • $\begingroup$ why is the first integration on the right hand side of the second line from below does not contain $dx$? $\endgroup$ – Smart Apr 18 at 13:46
  • $\begingroup$ The infinitesimal Euclidean arc length is just $\sqrt{dx^2 + dy^2}$, which you can "factor" the $dx$ outside and have an integral depending only on one parameter. $\endgroup$ – Dayton Apr 18 at 13:56
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Here one derivation starting from the definitions. Let $\tau:[a,b]\rightarrow R^2:t\rightarrow(t,f(t))$ a parametrization of $C$.

By definition the integral of a differential form $\rho$ is:

$\int_C \rho =\int_{[a,b]} \rho(\tau(t))[\tau'(t)]dt$

, where $\rho(a)[b]$ is the differential form at $a$ acting on vector $b$.

when $\rho=Fdx$, since $\tau'(t)=(1,f'(t))$, $dx[\tau'(t)]=1$ we have:

$\int_C F(x,y) dx =\int_{[a,b]} F(t,f(t))dt$

when $\rho=Fdy$, since $\tau'(t)=(1,f'(t))$, $dy[\tau'(t)]=f'(t)$ we have:

$\int_C F(x,y) dy =\int_{[a,b]} F(t,f(t))f'(t)dt$

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